Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}-\frac{-2}{9}-\frac{7}{5}\right)-\frac{5x}{2}\left(\frac{x}{5}-\frac{4}{5}\right)\)
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}+\frac{2}{9}-\frac{7}{5}\right)-\frac{5x}{2}.\frac{x-4}{5}\)
\(M=\frac{-2x}{3}+3x\left(\frac{15x+20-126}{90}\right)-\frac{5x^2-20x}{10}\)
\(M=\frac{-2x}{3}+3x.\frac{15x-106}{90}-\frac{5.\left(x^2-4x\right)}{10}\)
\(M=\frac{-2x}{3}+\frac{45x^2-318x}{90}-\frac{x^2-4x}{2}\)
Bài 1:
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
= \(\left(\frac{1}{5}-3\right)x^4y^3\)
= \(-\frac{14}{5}x^4y^3.\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
= \(\left(5-\frac{1}{4}\right)x^2y^5\)
= \(\frac{19}{4}x^2y^5.\)
Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.
Chúc bạn học tốt!
\(C=\frac{7}{9}x^3y^2\left(\frac{6}{11}axy^3\right)+\left(-5bx^2y^4\right)\left(\frac{-1}{2}axz\right)+ax\left(x^2y\right)^3\)
\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax\left(x^6y^3\right)\)
\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax^7y^3\)
\(D=\frac{\left(3x^4y^4\right)^2\left(\frac{6}{11}x^3y\right)\left(8x^{n-7}\right)\left(-2x^{7-n}\right)}{15x^3y^2\left(0,4ax^2y^2z^2\right)^2}\)
\(D=\frac{\left[3.\frac{6}{11}.8.\left(-2\right)\right]\left(x^8x^3x^{n-7}x^{7-n}\right)\left(y^8y\right)}{15.0,4.\left(x^3x^4\right)\left(y^2y^4\right)z^4a}\)
\(D=\frac{\frac{-188}{11}x^{24}y^9}{6x^7y^6z^4a}\)
a) \(x\ge\frac{2}{3}\Rightarrow3x-2\ge0\Rightarrow\left|3x-2\right|=3x-2\)
\(\Rightarrow P=\frac{1}{2}-\frac{1}{2}\left(x:\frac{1}{6}-\frac{1}{4}\right)-2\left(3x-2\right)\)
\(\Rightarrow P=\frac{1}{2}-\frac{1}{2}\left(6x-\frac{1}{4}\right)-6x+4\)
\(\Rightarrow P=\frac{4}{8}-3x+\frac{1}{8}-6x+\frac{32}{8}\)
\(\Rightarrow P=\frac{37}{8}-9x\)
b) \(x< \frac{2}{3}\Rightarrow3x-2< 0\Rightarrow\left|3x-2\right|=2-3x\)
\(\Rightarrow P=\frac{1}{2}-\frac{1}{2}\left(x:\frac{1}{6}-\frac{1}{4}\right)-2\left(2-3x\right)\)
\(\Rightarrow P=\frac{1}{2}-\frac{1}{2}\left(6x-\frac{1}{4}\right)-4+6x\)
\(\Rightarrow P=\frac{4}{8}-3x+\frac{1}{8}-\frac{32}{8}+6x\)
\(\Rightarrow P=\frac{-27}{8}+3x\)